# Delta function integral

Homework Helper
Evaluate:

$$\int^{\infty}_{-\infty} f(x)\delta(x-x_0)dx$$

Where

$$f(x)=ln(x+3), x_0=-2$$

Ordinarily, you would just evaluate $$f(x_0)$$, so it would be 0, but in this case, since $$f(x)$$ is $$-\infty$$ at $$x=-3$$, does that make a difference?

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But you have x0=-2 and f(x) is certainly not negative infinity there. What is f(-2)?

Homework Helper
Right, f(-2) = ln(1) = 0

I'm just wondering if for the formula:

$$\int_{-\infty}^{\infty}f(x)\delta (x-x_0) = f(x_0)$$

there need to be certain conditions on f(x), that would make a logarithmic function not suitable? My textbook was very vague in this area.

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Hurkyl
Staff Emeritus
Gold Member
You're missing the + and the - on your limits...

There are ways to define integration of generalized functions so that this integral makes sense for any (partial) function f that is well-behaved near x0. It might even be possible to make this integral well-defined when f is a generalized function that is well-behaved near x0!

HallsofIvy
Homework Helper
Right, f(-2) = ln(1) = 0

I'm just wondering if for the formula:

$$\int_{-\infty}^{\infty}f(x)\delta (x-x_0) = f(x_0)$$

there need to be certain conditions on f(x), that would make a logarithmic function not suitable? My textbook was very vague in this area.
No. For any function defined at x0,
$$\int_{-\infty}^\infty f(x)\delta (x)dx= f(x_0)$$

Hurkyl
Staff Emeritus
$$\int_{-\infty}^\infty f(x)\delta (x)dx= f(x_0)$$